loading解：任取一只元件，其寿命大于1500h的概率为P(X>1500)=??1000x21500dx?23
以Y记所取5只中寿命大于1500h的元件的数目，则Y服从二项分布
B(5,),故所求概率为
32P?Y?2??1?P?Y?0??P?Y?1??2??1?1?2??1??1?C?????C5????050514?3??3??232243
?3??3?
概率答案
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解：任取一只元件，其寿命大于1500h的概率为P(X>1500)=??1000x21500dx?23
以Y记所取5只中寿命大于1500h的元件的数目，则Y服从二项分布
B(5,),故所求概率为
32P?Y?2??1?P?Y?0??P?Y?1??2??1?1?2??1??1?C?????C5????050514?3??3??232243
?3??3?